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Undergraduate Journal of Mathematical Undergraduate Journal of Mathematical

Modeling: One + Two Modeling: One + Two

Volume 10 | 2020 Spring 2020 Article 2

2020

Prosthetic Leg Model Prosthetic Leg Model

Dang Nguyen University of South Florida

Advisors:

Arcadii Grinshpan, Mathematics and Statistics

Dmitri Voronine, Physics

Problem Suggested By: Dmitri Voronine

Field of Study for Problem Suggester: Physics

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Recommended Citation Recommended Citation Nguyen, Dang (2020) "Prosthetic Leg Model," Undergraduate Journal of Mathematical Modeling: One + Two: Vol. 10: Iss. 2, Article 2. DOI: https://doi.org/10.5038/2326-3652.10.2.4915 Available at: https://digitalcommons.usf.edu/ujmm/vol10/iss2/2

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Prosthetic Leg Model Prosthetic Leg Model

Abstract Abstract The main goal of this paper is to introduce an imitated prosthetic leg model by analyzing the applied forces. Even though the model is based on the idea of a prosthetic leg, it is also applicable to people without disabilities. The same concept of the model can be seen in the circus, where a person maintains a balanced state while on an incredible height without falling. When all the applying forces in the system are calculated, the design achieves the ideal state which allows it to function most effectively. One of the essential factors applied to the imitated prosthetic leg model, that causes a great impact on the human body’s movement, is torque. Torque is proportional to the spinning force that tends to cause rotation about a certain axis. Therefore, the paper also focuses on the idea of moment of inertia and center of mass to estimate torque. The density of different materials is also considered and compared in order to provide insights as to which would be the most effective.

Keywords Keywords prosthetic leg, torque, moment of inertia, center of mass

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1

PROBLEM STATEMENT

Based on the idea of how a prosthetic leg operates, we analyze and calculate forces that apply to

the new model in movement and, then, make the adjustment to its length and material for it to be

applied in real life. We also consider an environmental condition when the model travels through

a liquid.

MOTIVATION

Newtonian forces play a crucial role not only in science and engineering but also provide us with

countless applications on a daily basis. According to classical physics, suitable to describe the

macroscopic world, the motion of any object in the universe is not random but rather

quantifiable, calculable, and therefore predictable. As a result, the movement of objects and

humans can also be analyzed and calculated by considering the effects of the applied forces.

Therefore, it is essential to be able to understand the nature of the forces and what factors

influence the movement of the object. This paper examines one example which includes a

variety of forces: normal, gravitational, buoyancy, drag, and static friction, all of which affect a

certain rotating object. It also applies the idea of moment of inertia and center of mass in order to

illustrate precisely the rotational effect resulting from the aforementioned forces. All of them are

interpreted based on the classical mechanics and Newton’s second law of motion.

MATHEMATICAL DESCRIPTION AND SOLUTION APPROACH

A) ANALYZING FORCES

First of all, we need to analyze all the equations that are important in explaining how the model

functions. The main theme of this paper utilizes Newton’s second law of motion:

Nguyen: Prosthetic Leg Model

Produced by The Berkeley Electronic Press, 2020

2

∑ 𝐹 = 𝑀𝑎

This equation predicts the movement of the model under the condition when the forces applied

on it are unbalanced. We will take normal, gravitational, buoyancy, drag and static frictional

forces into consideration when calculating the applied forces, namely:

𝑁 = 𝑀𝑔𝑐𝑜𝑠(𝑥) (1)

𝐹𝑔 = 𝑀𝑔 (2)

𝐹𝐵 = 𝑀𝑓𝑙𝑢𝑖𝑑 ∗ 𝑔 (3)

𝐹𝐷 = 1

2𝜌𝑣2𝐶𝐷𝐴 (4)

𝑓𝑠 = 𝜇𝑠 ∗ 𝑁 (5)

Equations (1) and (2) represent the forces that we and most of the objects experience on a daily

basis, i.e. normal and gravitational forces. The variable “x” is the angle of incline since the

normal force only exerts an action perpendicular to the surface, “ 𝑔 ” is the gravitational

acceleration which approximately is given by 9.81 m/𝑠2 and M is the mass of the object itself.

The equation illustrates how we are constantly affected by the earth and the surface that we are

standing on. Equation (3) depicts the buoyancy force that is created by a fluid (such as an object

crosses a river), where 𝑀𝑓𝑙𝑢𝑖𝑑 is the mass of fluid that the object displaces and it can be presented

by:

𝜌 = 𝑀𝑓𝑙𝑢𝑖𝑑

𝑉 ,

where 𝜌 is the density of freshwater, V is the volume of water that the object displaces.

We consider freshwater for river, with density approximately equal to 1000 kg/𝑚3 and without

any viscosity. To calculate the drag force by the river, we apply equation (4) that still uses “𝜌”

for freshwater and where 𝐶𝐷 is the drag coefficient that depends on the material. Lastly, we have

the static friction. Equation (5) is included in order to prevent a user from slipping, and the static

coefficient depends on two factors: the surface that our model travels on and the material of the

model itself.

Undergraduate Journal of Mathematical Modeling: One + Two, Vol. 10, Iss. 2 [2020], Art. 2

https://digitalcommons.usf.edu/ujmm/vol10/iss2/2DOI: https://doi.org/10.5038/2326-3652.10.2.4915

3

Figure 1: Different forces apply on human body when using the model

Before we can analyze the forces affecting a moving object, which in this case is the human leg

combined with the designed model when a person is walking, it is important to understand the

mechanism of the human leg’s movement which is called the gait cycle. We will only consider

two main phases and analyze the applied forces: “Mid stance” for both legs in phase 1 and

“Terminal swing” for one leg in phase 2. Chakravarthy, Sultana, Srinath, and Kumar conducted a

research in 2017 that culminated in the paper “A mathematical model to determine the torque for

a prosthetic leg – Lagrangian equation”, in which this cycle is illustrated:

Figure 2: Gait cycle (Chakravarthy, Sultana, Srinath and Kumar, 2017, p.52)



4

1st Phase of the movement:

Figure 3: both legs in “Mid stance” position

2nd Phase of the movement:

Figure 4: One leg in “Mid stance” and one leg in “Terminal swing” position

These two phases represent the main movement when using the design. Note that we already

combine the human leg and the model as a “straight rod” having neglected the rotational motion

at the knee. Moreover, there are forces on the vertical direction such as normal, gravitational and

buoyancy forces which can be balanced out (acceleration in 𝑖̂ = 0). Because of that, we will shift

our attention to the forces in the horizontal direction such as drag and frictional forces

(Numerical value will be used in part C -Application- to make actual calculations for the

functional model).

Based on this idea, we can estimate the magnitude of the outcome forces on the model if there is

enough data. However, since the human leg is rather rotational motion than translational motion,

i.e. it rotates around the hip joint; we also need to take torque into account. The main objective is

to calculate all the forces and estimate the range for ideal torque so that a human can use the

model effectively, since the human hip joint can only exert in practice a certain amount of



5

torque. For the sake of simplicity, we only consider torque at the hip joint and neglect the

rotational motion in the knee by considering the prosthetic leg and real human leg combined as

one straight, circular “rod”:

𝜏 = 𝐹 ∗ 𝑟,

𝜏 = Ι ∗ 𝛼,

where “I” stands for the moment of inertia, which can be calculated by the following equation, in

which m is mass of the object and r is the distance from the mass to its axis of rotation:

Ι = ∑ 𝑚𝑖𝑟𝑖2𝑁

𝑖=1 .

This equation is mainly used for calculating the moment of inertia of the system where there are

more than one mass interconnected with each other. To be able to estimate the moment of inertia

for this model, we need to reform integration. This procedure consists of two parts: the “uniform

rod”-combination of the real human leg and the model, and the “base” which serves as a bridge

to connect the model with the surface. Rotational axis will be located at the hip joint and the

length of “uniform rod” is L:

Figure 5: Model illustrates the gravitational force and torque applied on human leg

Note: The model’s size and weight will be modified so the combination between human leg and

the model will get as close to the uniform rod as possible



6

We know that the density of the object, based on its dimensions, can be represented as:

1𝐷𝜆 = 𝑚

𝑙𝑒𝑛𝑔𝑡ℎ

2𝐷𝜎 = 𝑚

𝐴𝑟𝑒𝑎

3𝐷𝜌 = 𝑚

𝑉𝑜𝑙𝑢𝑚𝑒

B) CALCULATIONS

Part 1: Calculating the moment of inertia for the combination of the model and human leg

around the axis of rotation

Figure 6: Calculating the moment of inertia for the rod with an axis of rotation at one end

Note: The image is shown in 2 dimensions and the axis of rotation lines along y-axis for the

observation purpose but the result will be assumed to be the same.

We have: 1𝐷𝜆 = 𝑚

𝑙𝑒𝑛𝑔𝑡ℎ

Ι = ∫ 𝑟2𝑑𝑚 , where x can substitute for r, dm can be substituted by 𝜆𝑑𝑥

Ι = ∫ 𝑥2𝜆𝑑𝑥𝐿

0 = 𝜆 ∫ 𝑥2𝑑𝑥

𝐿

0

Ι = 𝜆𝑥3

3 𝑓𝑟𝑜𝑚 0 𝑡𝑜 𝐿

Ι𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 ℎ𝑢𝑚𝑎𝑛 𝑙𝑒𝑔 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑚𝑜𝑑𝑒𝑙 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑥𝑖𝑠 𝑎𝑡 ℎ𝑖𝑝 𝑗𝑜𝑖𝑛𝑡 = 𝑚

𝐿∗

𝐿3

3=

𝑚𝐿2

3



7

Part 2: Calculating the moment of inertia for the base

First case: The rectangular prism. When calculating the moment of inertia for the base, we

notice that the axis of rotation is in the different location, therefore we need to calculate the

moment of inertia that passes through the center of mass of the rectangular prism, and then we

apply the parallel axis theorem to shift the axis of rotation to the hip joint. In this specific case,

the moment of inertia is not affected whether the base is in 2D or 3D, therefore, we will integrate

the rectangle with the axis passing through its center (the center of mass of a symmetrical shape

is always at its center, including rectangle).

Figure 7: Calculating the moment of inertia for the base with rectangle prism shape

We have: h for height, b for base

2𝐷𝜎 = 𝑚

𝐴𝑟𝑒𝑎 𝑚𝑖 =

𝑚

ℎ𝑏

dA = dx*dy

Ι = ∬ 𝑚𝑟2𝑑𝐴 = ∫ 𝑑𝑦ℎ/2

−ℎ/2∫ 𝑑𝑥

𝑏

2

−𝑏

2

𝑚

ℎ𝑏 (𝑥2 + 𝑦2)

Ι = 𝑚

ℎ𝑏(∫

𝑥3

3𝑑𝑦 |

ℎ/2

−ℎ/2

𝑏/2−𝑏/2

+ ∫ 𝑦2𝑥 |ℎ/2

−ℎ/2

𝑏/2−𝑏/2

𝑑𝑦)

Ι = 𝑚

ℎ𝑏 (

𝑏3𝑦

3∗4 |

ℎ/2−ℎ/2

+ 𝑦3

3𝑏 |

ℎ/2−ℎ/2

)

Ι = 𝑚

ℎ𝑏 (

𝑏3ℎ

12 +

𝑏ℎ3

12)



8

Ι = 𝑚

12 (𝑏2 + ℎ2)

This is the moment of inertia for the rectangular prism whose rotational axis is right at its

centroid, therefore its Ι𝐶𝑜𝑀 is also equal to 𝑚

12 (𝑏2 + ℎ2). Now we apply the parallel axis

theorem to shift the axis:

Ι∣∣ = Ι𝐶𝑜𝑀 + 𝑚𝑑2

(Ι∣∣ is the new moment of inertia that has the rotational axis parallel to the moment of inertia at

the center of mass, and d is the distance between the two axes)

Figure 8: Calculating the moment of inertia for the rectangle prism with axis of rotation shifted

We will use L as the length of the combination of the human leg and the model:

Ι∣∣ = 𝑚

12 (𝑏2 + ℎ2) + 𝑚 ((

𝑏

2)

2

+ (ℎ2

2+ 𝐿)

2

)

= 𝑚𝑏2

12+

𝑚ℎ2

12+

𝑚𝑏2

4+ 𝑚 (

ℎ4

4+

2ℎ2

2𝐿 + 𝐿2)

= 𝑚𝑏2

12+

𝑚ℎ2

12+

𝑚𝑏2

4+

𝑚ℎ4

4+ 𝑚ℎ2𝐿 + 𝑚𝐿2

= 𝑚𝑏2

3+

𝑚ℎ2

12+

𝑚ℎ4

4+ 𝑚ℎ2𝐿 + 𝑚𝐿2

This is the moment of inertia for the base that has a rectangular prism shape whose axis of

rotation passes though the human’s hip joint. As a result, the moment of inertia for the whole

human leg, which includes the leg, the model and the base, will be:



9

I = 𝑚𝑏2

3+

𝑚ℎ2

12+

𝑚ℎ4

4+ 𝑚ℎ2𝐿 + 𝑚𝐿2 +

𝑚𝐿2

3

= 𝑚𝑏2

3+

𝑚ℎ2

12+

𝑚ℎ4

4+ 𝑚ℎ2𝐿 +

4𝑚𝐿2

3 (6)

Second case: The triangular prism. This case is more complicated, but its flexibility is higher

than the rectangular prism and therefore requires less material. We will find its moment of inertia

by applying the same steps as for the rectangular prism. Since for the triangle the center of mass

is unknown, we first apply the following equations to obtain it:

𝑋𝑐𝑚 = ∑ 𝑚𝑖𝑥𝑖

𝑛𝑖

∑ 𝑚𝑖𝑛𝑖

= 𝑚1𝑥1 + 𝑚2𝑥2 + ⋯ + 𝑚𝑛𝑥𝑛

𝑚1 + 𝑚2 + ⋯ + 𝑚𝑛

𝑌𝑐𝑚 = ∑ 𝑚𝑖𝑦𝑖

𝑛𝑖


= 𝑚1𝑦1 + 𝑚2𝑦2 + ⋯ + 𝑚𝑛𝑦𝑛

𝑚1 + 𝑚2 + ⋯ + 𝑚𝑛

𝑍𝑐𝑚 = ∑ 𝑚𝑖𝑧𝑖

𝑛𝑖


= 𝑚1𝑧1 + 𝑚2𝑧2 + ⋯ + 𝑚𝑛𝑧𝑛

𝑚1 + 𝑚2 + ⋯ + 𝑚𝑛

Instead of finding CoM for a 3D shape, the triangular prism, we can find CoM for the triangle

which is two-dimensional, since by reasons of symmetry we know, that the 3D CoM would lie

along a line that is perpendicular to the plane we find in two dimensions.

We know that:

𝑋𝑐𝑚 = ∫ 𝑥𝑑𝑚

∫ 𝑑𝑚 , 𝑌𝑐𝑚 =

∫ 𝑦𝑑𝑚

𝑑𝑚 ,

where 𝑑𝑚 is the small part of the total mass.

Calculating 𝑋𝑐𝑚

Figure 9: Calculating the center of mass of the triangle in x-axis



10

𝑋𝑐𝑚 = ∫ 𝑥𝑑𝑚

∫ 𝑑𝑚 =

1

𝑚 ∫ 𝑥𝑑𝑚

𝑏

0=

1

𝑚 ∫ 𝑥𝜎𝑦𝑑𝑥

𝑏

0

Since 𝑦 = ℎ𝑥

𝑏

𝑋𝑐𝑚 = 1

𝑚(

𝜎ℎ

𝑏) ∫ 𝑥2𝑑𝑥

𝑏

0=

𝜎ℎ𝑏3

3𝑚𝑏=

𝜎ℎ𝑏2

3𝑚

𝑋𝑐𝑚 = 𝑚

1

2𝑏ℎ

(ℎ𝑏2

3𝑚) =

2

3𝑏 (7)

Calculating 𝑌𝑐𝑚

Figure 10: Calculating the center of mass of the triangle in y-axis

𝑌𝑐𝑚 = ∫ 𝑦𝑑𝑚

𝑑𝑚=

1

𝑚 ∫ 𝑦𝑑𝑚

Substitute 𝑑𝑚 for 𝜎𝑑𝐴, where dA = (b-x)dy. We also know that 𝑥 =𝑏𝑦

ℎ, therefore

𝑌𝑐𝑚 = 1

𝑚 ∫ 𝑦𝜎𝑑𝐴 =

1

𝑚 ∫ 𝑦𝜎(𝑏 − 𝑥)𝑑𝑦

ℎ

0

𝑌𝑐𝑚 = 1

𝑚𝜎 ∫ 𝑦 (𝑏 −

𝑏𝑦

ℎ)

ℎ

0𝑑𝑦 =

1

𝑚𝜎 (

𝑏ℎ2

2−

𝑏ℎ3

3ℎ) =

1

𝑚𝜎

𝑏ℎ2

6

And since 𝜎 = 𝑚

𝐴𝑟𝑒𝑎=

𝑚1

2𝑏ℎ

𝑌𝑐𝑚 = 1

𝑚 (

𝑚1

2𝑏ℎ

)𝑏ℎ2

6=

ℎ

3 (8)

We can now obtain the coordinate of the CoM of the triangle based on its base and height:

(2𝑏

3 ,

ℎ

3 ). The axis of rotation passes through this point and also passes through the CoM of the



11

triangular prism and it is perpendicular to the object’s surface (which means it is in the direction

of z-plane).

The second problem of the triangular prism is its axis of rotation. To calculate the moment of

inertia in which the axis of rotation is right at the center of mass is complicated. Therefore, we

find moment of inertia about the line of rotation in figure 11, located at (𝑏

2 ,

ℎ

2), and then apply the

parallel axis theorem to find the moment of inertia about the center of mass, and apply the

parallel axis theorem for the second time to find the moment of inertia about the axis of rotation

located at the human’s hip joint.

Figure 11: Calculating the moment of inertia for the triangle prism with shifted axis of rotation

Now we integrate the triangular prism with the axis of rotation at (𝑏

2 ,

ℎ

2)

We have b for base, h for height and 𝑙 for length of the triangular prism.

Ι𝑧 = ∫(𝑥2 + 𝑦2)𝜌𝑑𝑣

= 𝜌𝑙 ∫ ∫ (𝑥2 + 𝑦2)𝑑𝑥𝑑𝑦ℎ−

ℎ𝑥

𝑏0

𝑏

0

Ι𝑧 = ∭(𝑥2 + 𝑦2)𝜌𝑑𝑣

= 𝜌𝑙 ∫ ∫ (𝑥2 + 𝑦2)𝑑𝑥𝑑𝑦ℎ−

ℎ𝑥

𝑏0

𝑏

0

Ι𝑦 = (𝑥2𝑦 + 𝑦3

3) 𝑓𝑟𝑜𝑚 0 𝑡𝑜 ℎ −

ℎ

𝑏𝑥



12

= (𝑏−

ℎ𝑥

𝑏)

3

3+ 𝑥2 (𝑏 −

ℎ𝑥

𝑏)

Ι = 𝜌𝑙 ∫ Ι𝑦𝑑𝑥𝑏

0=

𝑏𝑙𝜌 (8𝑏3−9𝑏2ℎ+4𝑏ℎ2−ℎ3)

12 , with 𝜌 =

𝑚

𝑉𝑜𝑙𝑢𝑚𝑛=

𝑚1

2𝑏ℎ𝑙

Ι = 𝑚 (8𝑏3−9𝑏2ℎ+4𝑏ℎ2−ℎ3)

6ℎ=

8𝑚𝑏3

6ℎ−

9𝑏2ℎ𝑚

6ℎ+

4𝑏ℎ2𝑚

6ℎ−

ℎ3𝑚

6ℎ

Ι = 4𝑏3𝑚

3ℎ−

3𝑏2𝑚

2+

2𝑏ℎ𝑚

3−

ℎ2𝑚

6

To find the moment of inertia with the axis of rotation at CoM, we have to flip the triangle to

sync the two triangles together. Therefore the CoM that we will use is at (𝑏

3,

ℎ

3) , and now we will

find the distance from it to the point (𝑏

2 ,

ℎ

2) to apply parallel axis theorem for the triangular

prism.

Figure 12: Calculating the moment of inertia for the triangle prism with axis of rotation located

at human’s hip joint.

Distance from (𝑏

3,

ℎ

3) to (

𝑏

2 ,

ℎ

2) = √((

𝑏

2−

𝑏

3)

2

+ (ℎ

2−

ℎ

3)

2)

= √(𝑏

6)

2

+ (ℎ

6)

2

= √𝑏2+ℎ2

36



13

= √𝑏2+ℎ2

6

Then we apply the parallel axis theorem: Ι∣∣ = Ι𝐶𝑜𝑀 + 𝑚𝑑2

Ι𝐶𝑜𝑀 = 4𝑏3𝑚

3ℎ−

3𝑏2𝑚

2+

2𝑏ℎ𝑚

3−

ℎ2𝑚

6−

𝑏2+ℎ2

36

Distance from CoM of the triangular prism to the human’s hip joint can be calculated by:

√(𝐿 + 2ℎ

3)

2

+ (𝑏

3)

2

We use L as the length of the combination of the human leg and the model. Now we apply the

parallel axis theorem again to obtain:

Ι∣∣ = ( 4𝑏3𝑚

3ℎ−

3𝑏2𝑚

2+

2𝑏ℎ𝑚

3−

ℎ2𝑚

6−

𝑏2+ℎ2

36 ) + 𝑚 ∗ [(𝐿 +

2ℎ

3)

2

+ (𝑏

3)

2

]

= 4𝑏3𝑚

3ℎ−

3𝑏2𝑚

2+

2𝑏ℎ𝑚

3−

ℎ2𝑚

6−

𝑏2+ℎ2

36+ 𝑚𝐿2 +

4𝑚𝐿ℎ

3+

4𝑚ℎ2

9+

𝑚𝑏2

9

For the triangular prism we calculate the moment of inertia for the combination of the model,

human leg and the base that has axis of rotation at human’s hip joint to be equal to:

4𝑏3𝑚

3ℎ−

3𝑏2𝑚

2+

2𝑏ℎ𝑚

3−

ℎ2𝑚

6−

𝑏2+ℎ2

36+ 𝑚𝐿2 +

4𝑚𝐿ℎ

3+

4𝑚ℎ2

9+

𝑚𝑏2

9+

𝑚𝐿2

3=

4𝑏3𝑚

3ℎ+

2𝑏ℎ𝑚

3−

𝑏2+ℎ2

36+

4𝑚𝐿ℎ

3+

4𝑚𝐿2

3−

25𝑏2𝑚

18+

5𝑚ℎ2

18 (9)

C) APPLICATION

In this part, we will take a closer look into the numerical data and calculate the actual range for

different values in order to create a functional imitated prosthetic leg model.

Part 1: Estimating the length of the combination between the human leg and the model, which

lead to measure its total weight



14

According to the previous research (see appendix), we use the following data:

- Average human weight: 75 kg

- Average mass for 1 human leg: 17-19% of total mass → ~13.5 kg

- Average human leg length: 81 cm = 0.81 m

The number varies according to country, race, age and gender; therefore, the value we use is just

an approximation to present the general formula but does not apply to any specific group or

individual.

Figure 13: Combination between the model and human’s leg

We have:

𝐿𝑡𝑜𝑡𝑎𝑙 = 𝐿𝑙𝑒𝑔 + 𝐿𝑚𝑜𝑑𝑒𝑙 = 0.81 𝑚 + 𝐿𝑚𝑜𝑑𝑒𝑙

𝑚𝑡𝑜𝑡𝑎𝑙 = 𝑚𝑙𝑒𝑔 + 𝑚𝑚𝑜𝑑𝑒𝑙 = 13.5 𝑘𝑔 + 𝑚𝑚𝑜𝑑𝑒𝑙

Correlation between length and mass of the model:

𝑚𝑚𝑜𝑑𝑒𝑙 = 𝜌 ∗ 𝑉𝑚𝑜𝑑𝑒𝑙

𝑉𝑚𝑜𝑑𝑒𝑙 = 𝜋𝑟2 ∗ 𝐿𝑚𝑜𝑑𝑒𝑙

Mass of the model depends on its radius, length and material



15

For the purpose of testing, we will use r = 0.1 m, 𝑙 = 1m and 3 different materials include wood,

rubber and steel.

Material 𝜌 (𝑘𝑔/𝑚3) 𝑉𝑚𝑜𝑑𝑒𝑙 (𝑚3) 𝐿𝑡𝑜𝑡𝑎𝑙 (𝑚) 𝑚𝑚𝑜𝑑𝑒𝑙(𝑘𝑔)

Wood 600-700 0.031 1.81 20.15

Rubber 1522 0.031 1.81 47.18

Steel 7750 0.031 1.81 240.25

Table 1: Density, volume and mass for 3 different materials

Part 2: Analyze how the forces apply to the model for 2 cases (on the ground and under the

water) and select the most effective material for the model.

Recall two fundamental torque equations:

𝜏𝑛𝑒𝑡 = Ι𝛼

𝜏𝑛𝑒𝑡 = 𝐹𝑟

First of all, it is important to know how much torque the average person can exert through the

hip joint, or in other words, how much force can make the quadriceps muscles handle without

resulting in any injury. However, the torque at the hip joint depends on a variety of factors such

as the speed that humans walk, other groups of muscles, the angle of knee flexion, etc.

Moreover, we are not looking for the threshold of torque that can be produced by the hip joint

but rather the average torque that a person exerts during the action of walking. Therefore, we will

use an approximate number from the previous research that illustrates a reasonable torque when

a human is walking. According to Reilley and Martens, the quadriceps muscle can take in more

than 600kg if the angle of knee flexion is over 130 degrees (graph provided in the Appendix).



16

𝜏 max = 6000 N

Set 𝜏′ as the torque that human hip joint creates, range from 0 to 6000N*m

Set 𝜏′′ as external torques created by the environment and applied on the leg in 𝑗̂ direction

𝜏𝑛𝑒𝑡 = 𝜏′ − 𝜏′′ ≥ 0

Set 𝜇𝑘1 as the coefficient of kinetic friction for the surface of the ground.

We start by analyze forces for 2 cases, on the ground and under the water by calculating 𝜏′′.

Ground: 𝜏′′ = 𝑓𝑘 = 𝜇𝑘1∗ 𝑁 , where 𝜇𝑘1

depends on the interaction between our model base

material and the surface. Now we will use concrete for the surface on the ground and determine

the material for the model base that minimizes the kinetic friction, which will lead to a reduction

in 𝜏′, the torque created by human hip joint.

Some examples for 𝜇𝑘1 between concrete and different materials:

- Wood: 0.62

- Rubber: 0.60 – 0.85 → 0.725

- Horseshoe (steel): 0.58

Material 𝐹𝑁𝑡𝑜𝑡𝑎𝑙 (N) 𝜇𝑘1

𝑓𝑘 (N)

Wood 565 0.62 350.3

Rubber 830 0.725 601.75

Steel 2722 0.58 1578.76

Table 2: Gravitational force and kinetic friction for 3 different materials

Note: 𝐹𝑁𝑡𝑜𝑡𝑎𝑙 = half body mass + model mass

Conclusion: Wood or Rubber serve as better materials for the model since their weights

and the frictional forces are less compared to Steel. Even though human hip joint is still



17

able to handle frictional force created by Steel (𝜏 max = 6000 N) but it requires more

energy to travel.

Water: 𝜏′′ = 1

2𝜌𝑣2𝐶𝐷𝐴 , where 𝐶𝐷 is the drag coefficient based on the shape of the object

immersed under water. For long cylinder, straight rod (which is the combination of the human

leg with the model), the drag coefficient equals 0.82. In the water, the frictional force will be

small enough to be neglected and we therefore shift our attention to the drag force.

After a closer look, the drag force does not depend on its material but it is rather based on

different factors as long as the volume of the object remains constant.

We have:

- Average velocity of the river: 0m/s → 3.1 m/s ~ 5.5 km/h

- Average human speed in the river: 2 km/h

- Velocity of the water relative to fluid: 5.5+2=7.5 km/h

- Since we set length of the model to be 1 meter (part 2-C), the length that under water is

0.8 meter

Cross-sectional Area of the model that immerses under water can be calculated by the length of

the part immersed times diameter of the rod:

0.8 * 0.1 * 2 = 0.16 𝑚2

𝜏′′ = 1

2𝜌𝑣2𝐶𝐷𝐴 =

1

2∗ 1000 ∗ 7.52 * 0.82 * 0.16 = 3690 N

Conclusion: Material of the model does not affect the drag force applied to it. Moreover,

the human hip joint can still overcome the drag force since the maximum torque it can

create is 6000 N, which is larger than 𝜏′′.

Part 3: Finding the required torque (𝜏′) that a human hip joint needs to create in order to achieve

certain speed. This part mainly focuses on the moment of inertia and its equation that we

calculated in part B – Calculating.



18

𝜏 = Ι𝛼

We utilize this equation in order to figure out how much torque should the human hip joint exert

to achieve the ideal speed. Since the leg travels in rotational motion, we have to deal with the

angular velocity (𝜔) and angular acceleration (𝛼). However, the angle between the two legs is

small enough due to the length of the model and the leg combined (1.81 meter), we can consider

𝑣 as 𝜔 and 𝑎 as 𝛼.

For a person, to start at rest (𝑣0 = 0) and achieve 5 km/h (𝑣𝑓 = 5) in 0.5 second, the average

acceleration is:

𝑎𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 𝑣𝑓 − 𝑣0

𝑡=

5

0.5= 10 𝑚/𝑠2

Calculating 𝜏′ with different values for b,h and m (with height = ½ base)

- Rectangular prism

𝑚𝑏2

3+

𝑚ℎ2

12+

𝑚ℎ4

4+

4𝑚𝐿4

3+ 𝑚ℎ2𝐿

L (m) m (kg) b (m) h (m) 𝜏′ (𝑁) = Ι𝛼

1.81 5 0.1 0.05 715.93

1.81 5 0.2 0.1 717.14

1.81 5 0.3 0.15 719.16

1.81 10 0.4 0.2 1443.99

1.81 10 0.5 0.25 1451.31

1.81 10 0.6 0.3 1460.29

Table 3: Mass, base and height of the rectangle prism to calculate the applied torque



19

- Triangular prism

4𝑏3𝑚

3ℎ+

2𝑏ℎ𝑚

3−

𝑏2 + ℎ2

36+

4𝑚𝐿ℎ

3+

4𝑚𝐿2

3−

25𝑏2𝑚

18+

5𝑚ℎ2

18

L (m) m (kg) b (m) h (m) 𝜏′ (𝑁) = Ι𝛼

1.81 5 0.1 0.05 225.28

1.81 5 0.2 0.1 233.82

1.81 5 0.3 0.15 244.04

1.81 10 0.4 0.2 511.91

1.81 10 0.5 0.25 539.07

1.81 10 0.6 0.3 569.59

Table 4: Mass, base and height of the rectangle prism to calculate the applied torque

Conclusion: When achieving ideal speed and acceleration, human hip joint can exert

more than enough force that it is required since the maximum torque it can possibly

create without cause any injuring is approximately 6000N.



20

DISCUSSION

The purpose of this experiment is to analyze the imitated prosthetic leg model. Through different

values of the input data, it shows the maximum torque that a human leg creates in order to

overcome forces in nature such as the static friction and drag forces, and even with the weight of

the model hanging on the leg. The quadriceps muscles can perform with only a small portion of

the force and energy to achieve the ideal velocity (5 km/h), which is the average walking speed

of a human. The drag force is the only force that stands out from the others, which requires

approximately 3690 N to achieve the ideal speed. Generally, the result is expected. In real life,

when we walk on the ground or under water, we feel resistant forces that have opposite

directions with the path we are moving; however, most of the time, we are able to overcome

those forces without any difficulty. Making the problem more complicated, we can have a more

detailed look at the forces coming into play, since attaching the model which includes not only

the rod but also the base requires us to calculate the forces in different places. Moreover, we

have to take torque into account by using a variety of methods such as the parallel axis theorem

or finding the moment of inertia for the 3D object. We are then able to determine the two general

equations for the rectangle and triangular prisms based on their masses, lengths, and heights.

Then, we test different values in order to compare them with each other. Surprisingly, the change

in the base and height does not affect much the torque that hip joint requires in order to exert the

force. However, the change in mass (which could be caused by the change in material and

followed by the increase in density) increases the required torque significantly. The result of this

paper solidifies the ideas about forces that have been studied for many years in the engineering

field, and also provides a different approach to analyzing of the force problems.



21

CONCLUSION AND RECOMMENDATIONS

The paper provides 3 major parts about illustrating the impact of forces on a moving object and

comparing the nature forces with torque that is created by the human hip joint. By not only

considering translational but also rotational motion, it analyzes a variety of forces and torques

and many other aspects that go along with it. Through introducing and examining the simple

prosthetic leg model, we have an idea of how forces work in nature. The purpose of parts A and

B is to provide the general equations regarding the many unknown values of the problem, such

as the mass of the human, dimensions of the model, the ideal velocity, etc. However, those

values are controllable, which means they do not depend on the environment but rather can be

adjusted by humans. On the other hand, in part C, we take a closer look into numerical values

and examine the functionality of the model. Eventually, the data proves that the torque created

by the human hip joint can overcome most of the forces in nature. The experiment can be

analyzed even further and give a more insightful outcome to future practitioners who are

interested in this project. They may take several other factors, hereby neglected, into account

such as calculating the torque in the human knee joint.



22

NOMENCLATURE

Symbol Meaning Units

𝐹𝐵

Buoyancy force N

𝐹𝐷

Drag force N

𝐶𝐷

Drag coefficient N/A

𝜌 Density 𝑘𝑔/𝑚3

𝜏 Torque N*m

𝜏 𝑚𝑎𝑥

Maximum torque that human hip

joint can create

N*m

𝜏′

Torque that hip joint creates N*m

𝜏′′ External torque N*m

𝜔 Angular velocity 𝜃/𝑠

𝛼

Angular acceleration 𝜃/𝑠2

𝑓𝑘

Kinetic friction N

𝜇𝑘1

Coefficient of kinetic friction on

concrete

N/A

L Length of the combination between

the model and the human leg

m

𝐶𝑜𝑀 Center of mass N/A



23

REFERENCES

Body Segment Data. (n.d.). Retrieved December 12, 2019, from

https://exrx.net/Kinesiology/Segments.

Chakravarthy, K., Sultana, M. S., Srinath, A., & Kumar, N. (2017). A mathematical model to

determine the torque for a prosthetic leg - Lagrangian equation. Retrieved December 12, 2019,

from https://acadpubl.eu/jsi/2017-116-5-7/articles/7/9.pdf.

DASSY professional workwear: HOW TO ... choose the right inside leg length. (n.d.). Retrieved

from https://www.dassy.eu/pt/About_us/News/the-right-inside-leg-length/.

Donald T. Reilly & Marc Martens (1972) Experimental Analysis of the Quadriceps Muscle

Force and Patello-Femoral Joint Reaction Force for Various Activities, Acta Orthopaedica

Scandinavica, 43:2, 126-137, DOI: 10.3109/17453677208991251

Drag coefficient. (2019, November 15). Retrieved December 12, 2019, from

https://en.wikipedia.org/wiki/Drag_coefficient.

Elert, G. (n.d.). Speed of a River. Retrieved December 12, 2019, from

https://hypertextbook.com/facts/2006/NervanaGaballa.shtml

Normal weight ranges: Body mass index (BMI). (n.d.). Retrieved December 12, 2019, from

https://www.cancer.org/cancer/cancer-causes/diet-physical-activity/body-weight-and-

cancer-risk/adult-bmi.html



https://acadpubl.eu/jsi/2017-116-5-7/articles/7/9.pdf

https://www.dassy.eu/pt/About_us/News/the-right-inside-leg-length/

https://doi.org/10.3109/17453677208991251

https://en.wikipedia.org/wiki/Drag_coefficient

24

APPENDIX

Figure 1: Amount of force is created by human hip joint based on degrees of knee flexion

Table 1: Normal weight ranges: Body mass index (BMI)



25

Table 2: Percentage of body parts compared to the whole body



26

Table 3: Short version for coefficient of kinetic friction between different materials

Table 4: Short version for drag coefficient based on different shapes



prosthetic leg model - digitalcommons.usf.edu

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